R3.2 Electron transfer reactionsIB Chemistry HL: Revision notes

Oxidation = electron loss / oxidation state increase / oxygen gain / hydrogen loss; reduction is the reverse. The oxidising agent is reduced; the reducing agent is oxidised. Balance half-equations in acid with H₂O, then H⁺, then e⁻: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.

Metals are oxidised more easily down a group; halogens are reduced less easily down group 17. Reactive metals + dilute HCl or H₂SO₄ give a salt and H₂.

Anode = oxidation, cathode = reduction in every cell. In a voltaic cell (spontaneous reaction → electrical energy) the anode is negative; electrons flow anode → cathode through the wire, and ions move through the salt bridge. Secondary cells are recharged by reversing the discharge reactions. In an electrolytic cell (electrical energy drives a non-spontaneous reaction) the anode is positive; ions carry the current in the electrolyte. Molten salts: metal at the cathode, non-metal at the anode.

Organic redox: 1° alcohol → aldehyde → carboxylic acid; 2° alcohol → ketone (acidified K₂Cr₂O₇). Reductions with LiAlH₄/NaBH₄ reverse these; H₂/Ni reduces alkynes → alkenes → alkanes.

A standard electrode potential, E⦵, is measured against the standard hydrogen electrode (0.00 V) at 298 K, 100 kPa and 1 mol dm⁻³. The more positive E⦵, the more easily the species on the left is reduced (stronger oxidising agent); the more negative, the stronger the reducing agent on the right.

E⦵cell = E⦵(cathode, reduction) − E⦵(anode, oxidation). Do not multiply E⦵ by stoichiometric coefficients. A positive E⦵cell means the reaction is spontaneous as written; if it is negative, the reverse reaction is spontaneous.

ΔG⦵ = −nFE⦵cell, where n is the number of moles of electrons transferred in the equation as written and F = 96 500 C mol⁻¹. A positive E⦵cell gives a negative ΔG⦵ (spontaneous). Example: MnO₄⁻ + 8H⁺ + 5Fe²⁺: E⦵cell = 1.51 − 0.77 = +0.74 V; ΔG⦵ = −5 × 96 500 × 0.74 = −357 kJ mol⁻¹.

Water competes with the ions. At the cathode the species with the more positive (less negative) E⦵ is reduced: reactive-metal ions (Na⁺, K⁺, Mg²⁺) stay in solution and water gives H₂: 2H₂O + 2e⁻ → H₂ + 2OH⁻. Cu²⁺ and Ag⁺ are reduced to the metal. At the anode the reduced form in the half-equation with the less positive E⦵ is oxidised: usually water, 2H₂O → O₂ + 4H⁺ + 4e⁻.

Concentration: in concentrated NaCl, Cl⁻ is oxidised to Cl₂ (E⦵ values 1.23 and 1.36 V are close); dilute NaCl gives mainly O₂. Electrode: in CuSO₄(aq), a copper anode dissolves (Cu → Cu²⁺ + 2e⁻) while an inert graphite anode gives O₂. Electrolysis of water (acidified) gives H₂ and O₂ in a 2 : 1 volume ratio.

Electroplating coats an object with a thin metal layer. The object is the cathode, the electrolyte contains ions of the plating metal, and the anode is usually the plating metal itself. Silver plating: cathode Ag⁺ + e⁻ → Ag; anode Ag → Ag⁺ + e⁻. The electrolyte concentration stays constant because metal dissolves from the anode as fast as it is deposited.